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Chris Gerig
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On the de Rham cohomology of 1-fromsforms in cotangent bundle.

edited grammar, changed "cohomology group" to "cohomology class"
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Spiro Karigiannis
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We know that a cotangent bundle $T^\star M$ has a canonical symplectic form and $M$ is a naturenatural Lagrangian submanifold of it. A well known result is that any submanifold $X=\{(p,f(p)): p\in M\}$, where $f$ is a closed one form is Lagrangian. Denote by $[f]$ the de Rham cohomology groupclass of $f$. Assume that we flow $X$ in a Hamiltonian direction to $Y$, then $Y$ will be a Lagrangian submanifold of $T^\star M$. My question is can we write it as $Y={(p,g(p)): p\in M}$ for some closed one form $g$.? If so, do we have $ [g]=[f]?$ Thanks in advance!

We know that a cotangent bundle $T^\star M$ has a canonical symplectic form and $M$ is a nature Lagrangian submanifold of it. A well known result is that any submanifold $X=\{(p,f(p)): p\in M\}$, where $f$ is a closed one form is Lagrangian. Denote by $[f]$ the de Rham cohomology group of $f$. Assume that we flow $X$ in a Hamiltonian direction to $Y$, then $Y$ will be a Lagrangian submanifold of $T^\star M$. My question is can we write it as $Y={(p,g(p)): p\in M}$ for some closed one form $g$.? If so, do we have $ [g]=[f]?$ Thanks in advance!

We know that a cotangent bundle $T^\star M$ has a canonical symplectic form and $M$ is a natural Lagrangian submanifold of it. A well known result is that any submanifold $X=\{(p,f(p)): p\in M\}$, where $f$ is a closed one form is Lagrangian. Denote by $[f]$ the de Rham cohomology class of $f$. Assume that we flow $X$ in a Hamiltonian direction to $Y$, then $Y$ will be a Lagrangian submanifold of $T^\star M$. My question is can we write it as $Y={(p,g(p)): p\in M}$ for some closed one form $g$.? If so, do we have $ [g]=[f]?$ Thanks in advance!

I Edited the question according to the comments of Serge and Tagging. Thanks them!
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Mathboy
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On the Hodge homologyde Rham cohomology of 1-froms in cotangent bundle.

We know that a cotangent bundle $T^\star M$ has a canonical symplectic form and $M$ is a nature Lagrangian submanifold of it. A well known result is that any submanifold $X=\{(p,f(p)): p\in M\}$, where $f$ is a closed one form is Lagrangian. Denote by $[f]$ the Hodge homologyde Rham cohomology group of $f$. Assume that we flow $X$ in a Hamiltonian direction to $Y$, then $Y$ will be a Lagrangian submanifold of $T^\star M$. My question is can we write it as $Y={(p,g(p)): p\in M}$ for some closed one form $g$.? If so, do we have $ [g]=[f]?$ Thanks in advance! Edit: I edit the question according to Serge's comments.

On the Hodge homology of 1-froms in cotangent bundle.

We know that a cotangent bundle $T^\star M$ has a canonical symplectic form and $M$ is a nature Lagrangian submanifold of it. A well known result is that any submanifold $X=\{(p,f(p)): p\in M\}$, where $f$ is a closed one form is Lagrangian. Denote by $[f]$ the Hodge homology group of $f$. Assume that we flow $X$ in a Hamiltonian direction to $Y$, then $Y$ will be a Lagrangian submanifold of $T^\star M$. My question is can we write it as $Y={(p,g(p)): p\in M}$ for some closed one form $g$.? If so, do we have $ [g]=[f]?$ Thanks in advance! Edit: I edit the question according to Serge's comments.

On the de Rham cohomology of 1-froms in cotangent bundle.

We know that a cotangent bundle $T^\star M$ has a canonical symplectic form and $M$ is a nature Lagrangian submanifold of it. A well known result is that any submanifold $X=\{(p,f(p)): p\in M\}$, where $f$ is a closed one form is Lagrangian. Denote by $[f]$ the de Rham cohomology group of $f$. Assume that we flow $X$ in a Hamiltonian direction to $Y$, then $Y$ will be a Lagrangian submanifold of $T^\star M$. My question is can we write it as $Y={(p,g(p)): p\in M}$ for some closed one form $g$.? If so, do we have $ [g]=[f]?$ Thanks in advance!

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Mathboy
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Mathboy
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Mathboy
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